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Question: Assume that two liquids in a U-shape tube are water and oil. Compute the density of oil if water sta...

Assume that two liquids in a U-shape tube are water and oil. Compute the density of oil if water stands 19cm19cm above the interface and oil stands 24cm24cm above the interface.

Explanation

Solution

Pressure of a fluid at any point will be the same throughout its volume if we neglect its weight. But the weight of a fluid is not negligible so it varies according to depth of a point. The more is the depth, the more will be the pressure at that layer of fluid. The pressure at the same level of the liquid is constant at any point on that level irrespective of the shape of the container.

Formula used:
P=P0+hρgP = {P_0} + h\rho g
Where, PP is the pressure at depth hh of the fluid
P0{P_0} is the pressure at the surface of the fluid (atmospheric pressure)
hh is the depth of the liquid
ρ\rho is the density of the liquid
gg is acceleration due to gravity

Complete step by step answer:
Let’s consider a point AA at a depth of hh from the surface of the fluid. The pressure exerted at any point of the same depth is represented by the formula
P=P0+hρgP = {P_0} + h\rho g
We are given the question that, water stands 19cm19cm above the interface and oil stands 24cm24cm above the interface.
Let’s look at the following diagram.

From the above figure, as points A and B are on the same level, pressure is the same at both the points.
Thus we can write, PA=PB{P_A} = {P_B} ……………………. (1)
Where, PA{P_A} is the pressure at point A and PB{P_B} is the pressure at point B.
Point A is at a depth h1=19cm{h_1} = 19cm from the surface and point B is at a depth of h2=24cm{h_2} = 24cm from the surface.
Let’s consider the densities of water and oil as ρ1{\rho _1} and ρ2{\rho _2} respectively.
Now substituting the corresponding values of variables in the pressure-depth formula, we get,
For water, at point A,
PA=P0+h1ρ1g{P_A} = {P_0} + {h_1}{\rho _1}g ………. (2)
For oil, at point B,
PB=P0+h1ρ1g{P_B} = {P_0} + {h_1}{\rho _1}g ………. (3)
From equation (1),
PA=PB{P_A} = {P_B}
Putting the values of PA{P_A} and PB{P_B} in this equation we get,
P0+h1ρ1g=P0+h2ρ2g\Rightarrow {P_0} + {h_1}{\rho _1}g = {P_0} + {h_2}{\rho _2}g
Cancelling P0{P_0} from both sides, we get,
h1ρ1g=h2ρ2g\Rightarrow {h_1}{\rho _1}g = {h_2}{\rho _2}g
Dividing by gg in both sides we get,
h1ρ1=h2ρ2\Rightarrow {h_1}{\rho _1} = {h_2}{\rho _2} ……….. (4)
The density of water in C.G.S units is 1gcm31gc{m^{ - 3}}
Now substituting the values of h1{h_1}, ρ1{\rho _1} and h2{h_2} in equation (4), we get,
19×1=24×ρ2\Rightarrow 19 \times 1 = 24 \times {\rho _2}
ρ2=1924=0.791gcm3\Rightarrow {\rho _2} = \dfrac{{19}}{{24}} = 0.791gc{m^{ - 3}}

Thus the density of oil in C.G.S units is 0.791gcm30.791gc{m^{ - 3}}.

Notes: Water is taken as a universal measure of relative density of a fluid. The density of water at 4C4^\circ C is 1gcm31gc{m^{ - 3}} in C.G.S units and 1000kgm31000kg{m^{ - 3}} in S.I units. Interface means the level where the fluids meet, which in this case is point B. the formula used is the pressure-depth formula because pressure is not constant throughout the fluid, but it is different at different depths.